jmc

We look for a factorization of $x^4-4x-1$ of the form $(x^2+ax+b)(x^2+cx+d).$ Thus, $$x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd=x^4-4x-1.$$Matching coefficients, we get $$\begin{array}{rl}a+c& =0,\\ ac+b+d& =0,\\ ad+bc& =-4,\\ bd& =-1.\end{array}$$From the first equation, $c=-a.$ Substituting, we get $$\begin{array}{rl}-a^2+b+d& =0,\\ ad-ab& =-4,\\ bd& =-1.\end{array}$$Then $b+d=a^2$ and $b-d=\frac{4}{a},$ so $b=\frac{a^3+4}{2a}$ and $d=\frac{a^3-4}{2a}.$ Hence, $$\frac{(a^3+4)(a^3-4)}{4a^2}=-1.$$This simplifies to $a^6+4a^2-16=0.$ This factors as $$(a^2-2)(a^4+2a^2+8)=0,$$so we can take $a=\sqrt{2}.$ Then $b=1+\sqrt{2},$ $c=-\sqrt{2},$ and $d=1-\sqrt{2},$ so $$x^4-4x-1=(x^2+x\sqrt{2}+1+\sqrt{2})(x^2-x\sqrt{2}+1-\sqrt{2}).$$Checking the discriminants, we find that only the second quadratic factor has real roots, so the sum of the real roots is $\boxed{\sqrt{2}}.$